Question

The average amount of milligrams (mg) of sodium in the water in a certain neighbourhood in Winnipeg is 60 mg/l, and the standard deviation is 10 mg/l. Assume the variable is normally distributed.
(a) If a sample of water is selected form the neighbourhood, what is the probability that the sodium content will be more than 70 mg/l?
(b) If 10 samples of the water from the neighbourhood is selected, what is the probability that the mean of the sample will be larger than 70 mg/l?
(c) Why is the probability for part (a) greater than that for part (b)?

Answer 1

$\mu= 60, \sigma=10$

(a) P(X>70)

X is normally distributed with mean 60 and standard deviation 10.

$P(X>70)=1-P(X\le70)$

$Z-value =\frac{X-\mu}{\sigma}$

$=\frac{70-60}{10}=1$. From Z table or =NORM.DIST(70,60,10,TRUE),

$P(Z\le1)= 0.841345$

Thus, $P(X>70)=1-0.841345=0.158655$

(b). $P(\bar{X}>70)$

$\bar{X}$ Is normally distributed with mean 60 and variance $\frac{10}{\sqrt10}$

$P(\bar{X}>70)=1-P(\bar{X}\le70)$

$Z-value =\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{n}}}$

$=\frac{70-60}{\frac{10}{\sqrt{10}}}=3.162278$

From Z table or =NORM.DIST(70,60,10/10^0.5,TRUE),$P(\bar{X}\le70)=$ 0.999217

Thus, $P(\bar{X}>70)=1-0.999217=0.000783$

(c). As the sample size increases, the sample mean tends to be closer to the population mean, and standard deviation decreases as the sample size increases. Part (a) has a sample size of 1 and the standard deviation is 10, while part (b) has a sample size of 10 leading to the standard deviation of$\frac{10}{\sqrt{10}}$ which is smaller than the standard deviation in part (a). Therefore, part (a) has a greater probability since the distribution around X is wider than the distribution around $\bar{X}$ in part (b).